Question

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "admin_panel";



$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT *
  FROM my_chart_data
  ORDER BY category ASC";
$result = $conn->query($sql);
echo("<SCRIPT LANGUAHE='JavaScript'>
var chart;


AmCharts.ready(function() {


  var chartData = AmCharts.loadJSON('data.php');


  chart = new AmCharts.AmSerialChart();
  chart.dataProvider = chartData;
  chart.categoryField = 'category';
  chart.dataDateFormat = 'YYYY-MM-DD';



  var graph1 = new AmCharts.AmGraph();
  graph1.valueField = 'value1';
  graph1.bullet = 'round';
  graph1.bulletBorderColor = "#FFFFFF";
  graph1.bulletBorderThickness = 2;
  graph1.lineThickness = 2;
  graph1.lineAlpha = 0.5;
  chart.addGraph(graph1);

  var graph2 = new AmCharts.AmGraph();
  graph2.valueField = 'value2';
  graph2.bullet = 'round';
  graph2.bulletBorderColor = "#FFFFFF";
  graph2.bulletBorderThickness = 2;
  graph2.lineThickness = 2;
  graph2.lineAlpha = 0.5;
  chart.addGraph(graph2);


  chart.categoryAxis.parseDates = true;


  chart.write("chartdiv");
</SCRIPT>");
        exit();
?>

我使用phpmyadmin制作了草图，graph1中存在语法错误。我不知道怎么解决这个问题，如果我错了，有人可以检查代码并更正此代码。

这是错误，

解析错误：语法错误，意外＆＃39; graph1＆＃39; （T_STRING）

Answer 1

您需要在作为参数传递给echo()的字符串中转义双引号。请查看以下更改：

-  graph1.bulletBorderColor = "#FFFFFF";
+  graph1.bulletBorderColor = \"#FFFFFF\";

-  graph2.bulletBorderColor = "#FFFFFF";
+  graph2.bulletBorderColor = \"#FFFFFF\";

-  chart.write("chartdiv");
+  chart.write(\"chartdiv\");

数据使用phyMyAdmin检索图表图表中的错误

1 个答案: